Today I have done leetcode's November LeetCoding Challenge with cpp.

November LeetCoding Challenge 13

Description

Reverse Words in a String

Given an input string s, reverse the order of the words.

A word is defined as a sequence of non-space characters. The words in s will be separated by at least one space.

Return a string of the words in reverse order concatenated by a single space.

Note that s may contain leading or trailing spaces or multiple spaces between two words. The returned string should only have a single space separating the words. Do not include any extra spaces.

Example 1:

Input: s = "the sky is blue"
Output: "blue is sky the"

Example 2:

Input: s = "  hello world  "
Output: "world hello"
Explanation: Your reversed string should not contain leading or trailing spaces.

Example 3:

Input: s = "a good   example"
Output: "example good a"
Explanation: You need to reduce multiple spaces between two words to a single space in the reversed string.

Constraints:

• 1 <= s.length <= 104
• s contains English letters (upper-case and lower-case), digits, and spaces ' '.
• There is at least one word in s.

Follow-up: If the string data type is mutable in your language, can you solve it in-place with O(1) extra space?

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class Solution {
public:
  string reverseWords(string s) {
    reverse(s.begin(), s.end());
    int len = s.length();
    bool hasSpace = true;
    int newLength = 0;
    for(int i = 0; i < len; ++i) {
      if(s[i] != ' ' || !hasSpace) {
        if(s[i] == ' ') hasSpace = true;
        else hasSpace = false;
        s[newLength] = s[i];
        newLength += 1;
      }
    }
    if(!newLength) return "";
    if(s[newLength-1] == ' ') newLength -= 1;
    s.resize(newLength);
    len = newLength;
    int begin = -1, end = -1;
    for(int i = 0; i < len; ++i) {
      if(s[i] != ' ') {
        if(begin == -1) begin = i;
        end = i+1;
      } else {
        if(begin != -1 && begin != end-1) {
          reverse(s.begin() + begin, s.begin() + end);
        }
        begin = -1;
        end = -1;
      }
    }
    if(begin != -1 && begin != end-1) reverse(s.begin() + begin, s.begin() + end);
    return move(s);
  }
};

// Accepted
// 58/58 cases passed (4 ms)
// Your runtime beats 91.32 % of cpp submissions
// Your memory usage beats 83.99 % of cpp submissions (7.1 MB)